Mechanics of Materials: cram pack

Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.

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1Formulas and methods

Axial stress & strain

Normal stress: σ = F/A
Strain: ε = ΔL/L
Hooke’s law: σ = E·ε
Elongation: ΔL = FL/(EA)
Thermal strain: ε_T = α·ΔT

Torsion of shafts

Shear stress: τ = T·r/J
Solid shaft: J = πd⁴/32
Hollow shaft: J = π(D⁴ − d⁴)/32
Angle of twist: φ = TL/(GJ)
Power: P = T·ω = 2πnT/60

Method

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

Shear force & bending moment

Shear and moment: V = ΣF on one side, M = ΣF·d on one side
Bending stress: σ = M·y/I
Rectangle: I = bh³/12
Section modulus: W = I/y_max
Simply supported, uniform load: M_max = wL²/8

Method

Find the support reactions from equilibrium of the whole beam.
Cut the beam at the section you need and keep one side.
Shear force: sum of the vertical forces on that side. Bending moment: sum of their moments about the cut.
For the maximum moment, find where the shear force is zero.

Centroid & second moment of area

Centroid: ȳ = ΣA_iy_i/ΣA_i
Rectangle: I = bh³/12
Parallel-axis theorem: I = I_c + A·d²
Composite section: I = Σ(I_c,i + A_id_i²)

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).

Beam deflection

Point load at midspan: δ = PL³/(48EI)
Uniform load, simply supported: δ = 5wL⁴/(384EI)
Cantilever, end load: δ = PL³/(3EI)
Cantilever, uniform load: δ = wL⁴/(8EI)
Superposition: δ = δ₁ + δ₂

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

2Practice problems

1.Axial stress & strainMedium

Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).

P = 80 kNd = 20 mmL = 3 mE = 100 GPa

ε = µε

2.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 50 kNsegment 1: Ø25 mm, 2 m, brass (E = 100 GPa)segment 2: Ø12 mm, 2 m, copper (E = 117 GPa)

σ_max = MPa

3.Torsion of shaftsMedium

A solid brass shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.

T = 500 N·md = 30 mmL = 2 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft transmits 10 kW at 750 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.

P = 10 kWN = 750 rpmτ_allow = 60 MPaθ_allow = 1°/mG = 39 GPa

d = mm

5.Shear force & bending momentMedium

Find the bending moment M_x at the section x = 9 m from support A. Use sagging moments positive and upward shear on the left face positive.

L = 10 mw = 5 kN/m from 0 to 10 mx = 9 m

M_x = kN·m

6.Shear force & bending momentHard

The simply supported beam carries a uniformly distributed load over part of its span. Find the maximum bending moment M_max in the beam.

L = 10 mw = 12 kN/m from 2 to 10 m

M_max = kN·m

7.Centroid & second moment of areaMedium

A circular tube has outside diameter D = 100 mm and inside diameter d = 80 mm. Find its radius of gyration r about the horizontal centroidal axis.

D = 100 mmd = 80 mm

r = mm

8.Centroid & second moment of areaHard

A welded I-section has a top flange 100 × 20 mm, a web 290 mm high and 12 mm thick, and a bottom flange 220 × 12 mm (width × thickness). Find its second moment of area I_x about the horizontal axis x through its centroid.

top flange: 100 × 20 mmweb: 12 × 290 mmbottom flange: 220 × 12 mm

I_x = × 10⁶ mm⁴

9.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 110 MPa, σᵧ = 110 MPa and τₓᵧ = −20 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 110 MPaσ_y = 110 MPaτ_xy = −20 MPaσ_z = 0

τ_abs = MPa

10.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 10 MPa, σᵧ = 80 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 25° counter-clockwise from the x-axis.

σ_x = 10 MPaσ_y = 80 MPaτ_xy = −60 MPaθ = 25°

σ_θ = MPa

11.Beam deflectionMedium

A cantilever carries a uniformly distributed load w over its whole length and a point load P at its free end. Using superposition, find the deflection δ at the free end.

P = 21 kNw = 8 kN/mL = 2.5 mE = 200 GPa (steel)I = 12 × 10⁶ mm⁴

δ = mm

12.Beam deflectionHard

A simply supported timber beam of rectangular section, 125 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth h_req of the section.

w = 3 kN/mL = 3.5 mE = 11 GPa (timber)b = 125 mm

h_req = mm

13.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 3 mE = 70 GPa (aluminium)b × h = 20 × 100 mm

P_cr = kN

14.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 25 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 25 kNn = 2.5L = 1.5 mE = 200 GPa (steel)

d = mm

3Solutions

1.Axial stress & strain

Answer: ε = 2546 µε

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·20²/4 = 314.2 mm²

σ = P/A = 80·1000/314.2 = 254.6 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 254.6/100000 = 0.002546 = 2546 µε

Answer

ε = 2546 µε

2.Axial stress & strain

Answer: σ_max = 442.1 MPa

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A₁ = π·25²/4 = 490.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 101.9 MPa

σ₂ = 442.1 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 442.1 MPa

Answer

σ_max = 442.1 MPa

3.Torsion of shafts

Answer: φ = 18.47 °

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.

Polar second moment of area

J = π·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·2000/(39000·79522) = 0.3224 rad

φ = 0.3224·180/π = 18.47°

Answer

φ = 18.47°

4.Torsion of shafts

Answer: d = 37.15 mm

Torque

ω = 2πN/60 = 2π·750/60 = 78.54 rad/s

T = P/ω = 10000/78.54 = 127.3 N·m

Diameter for strength

T = 127.3 N·m = 127324 N·mm

d_τ = ∛(16·127324/(π·60)) = 22.11 mm

Diameter for stiffness

θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm

G = 39 GPa = 39000 MPa

d_φ = (32·127324/(π·39000·1.745 × 10⁻⁵))^1/4 = 37.15 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(22.11, 37.15) = 37.15 mm

Answer

d = 37.15 mm

Here stiffness governs.

5.Shear force & bending moment

Answer: M_x = 22.5 kN·m

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 5·10 = 50 kN, acting at x̄ = 5 m

ΣM_A = 0: R_B·10 = 5·10·5

R_B = 250/10 = 25 kN

ΣF_y = 0: R_A = 50 − 25 = 25 kN

Cut at x = 9 m and keep the left part

Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.

M = R_A·x − (each load left of the cut × its distance to the cut)

M_x = 25·9 − 5·9·4.5 = 22.5 kN·m

Answer

M_x = 22.5 kN·m

6.Shear force & bending moment

Answer: M_max = 138.2 kN·m

Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut. Maximum moment: since dM/dx = V, the bending moment has its peak where the shear force passes through zero, either at a point load or inside a distributed load.

Support reactions

Equilibrium of the whole beam.

UDL resultant: W = w·ℓ = 12·8 = 96 kN, acting at x̄ = 6 m

ΣM_A = 0: R_B·10 = 12·8·6

R_B = 576/10 = 57.6 kN

ΣF_y = 0: R_A = 96 − 57.6 = 38.4 kN

Zero shear

The shear force starts at R_A, stays constant over any unloaded part and falls with slope w inside the load.

R_A − w·(x₀ − 2) = 0

x₀ = 2 + 38.4/12 = 5.2 m

Maximum moment

M_max = R_A·x₀ − w·(x₀ − 2)²/2 = 38.4·5.2 − 12·3.2²/2 = 138.2 kN·m

Answer

M_max = 138.2 kN·m

7.Centroid & second moment of area

Answer: r = 32.02 mm

Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = I_c + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre. The radius of gyration r = √(I/A) is the distance at which the whole area would give the same I; it is used in column buckling.

Net area and second moment of area

A = π(D² − d²)/4 = 2827 mm²

I = π(D⁴ − d⁴)/64 = 2.898 × 10⁶ mm⁴

Radius of gyration

r = √(I/A) = √(2.898 × 10⁶/2827) = 32.02 mm

Answer

r = 32.02 mm

8.Centroid & second moment of area

Answer: I_x = 131.8 × 10⁶ mm⁴

Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.

Split into three rectangles

Measure every y from the bottom edge.

top flange: A = 2000 mm², y = 312 mm

web: A = 3480 mm², y = 157 mm

bottom flange: A = 2640 mm², y = 6 mm

Centroid

ȳ = (2000·312 + 3480·157 + 2640·6)/8120 = 146.1 mm

Parallel-axis theorem

top flange: 100·20³/12 + 2000·165.9² = 5.512 × 10⁷ mm⁴

web: 12·290³/12 + 3480·10.92² = 2.48 × 10⁷ mm⁴

bottom flange: 220·12³/12 + 2640·140.1² = 5.184 × 10⁷ mm⁴

I_x = 5.512 × 10⁷ + 2.48 × 10⁷ + 5.184 × 10⁷

I_x = 1.318 × 10⁸ mm⁴ = 131.8 × 10⁶ mm⁴

Answer

I_x = 131.8 × 10⁶ mm⁴

9.Mohr’s circle & stress transformation

Answer: τ_abs = 65 MPa

Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 110 MPa, R = √(0² + (−20)²) = 20 MPa

σ₁ = 130 MPa, σ₂ = 90 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(40, 130, 90)/2 = 65 MPa

Answer

τ_abs = 65 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

10.Mohr’s circle & stress transformation

Answer: σ_θ = −23.46 MPa

Transformation equation

σ_θ = 45 + −35·0.6428 + −60·0.766 = −23.46 MPa

Answer

σ_θ = −23.46 MPa

11.Beam deflection

Answer: δ = 61.85 mm

Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

P = 21000 N, w = 8 N/mm, L = 2500 mm

EI = 2.4 × 10¹² N·mm²

Each load on its own

δ_P = PL³/(3EI) = 21000·2500³/(3·2.4 × 10¹²) = 45.57 mm

δ_w = wL⁴/(8EI) = 8·2500⁴/(8·2.4 × 10¹²) = 16.28 mm

Add them

δ = 45.57 + 16.28 = 61.85 mm

Answer

δ = 61.85 mm

12.Beam deflection

Answer: h_req = 173.9 mm

Consistent units

Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.

w = 3 N/mm, L = 3500 mm, E = 11000 N/mm²

Deflection limit

δ_allow = L/360 = 3500/360 = 9.722 mm

Required second moment of area

I_req = 5wL⁴/(384·E·δ_allow) = 5·3·3500⁴/(384·11000·9.722) = 5.481 × 10⁷ mm⁴

Required depth

bh³/12 = I_req → h = (12·5.481 × 10⁷/125)^1/3 = 173.9 mm

Answer

h_req = 173.9 mm

In practice, choose the next standard depth above this value.

13.Column buckling

Answer: P_cr = 10.44 kN

Euler buckling: a slender column becomes unstable at P_cr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.

Smallest second moment of area

A rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 100·20³/12 = 66667 mm⁴

Effective length

KL = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·66667/2100² = 10444 N = 10.44 kN

Answer

P_cr = 10.44 kN

14.Column buckling

Answer: d = 49.09 mm

Required critical load

P_cr = n·P = 2.5·25000 = 62500 N

KL = 2·1500 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·62500·3000²/(π³·200000))^(1/4) = 49.09 mm

Answer

d = 49.09 mm

Check: the slenderness KL/r is 244.5, so the column is slender and Euler applies.

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